Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 47 / 3 / b

Use the ratio-of-uniforms method onWriting $x=v/u$ gives $u\le e^{-x^2/(4{\sigma }^2)}$ and $v=ux$. Consequentlybecause the derivative vanishes at $x=\sigma \sqrt{2}$. Thus the normal ratio-of-uniforms envelope is the rectangle $0<u<1$, $|v|<\sigma \sqrt{2/e}$.

Generate independent $U_1,U_2$ with uniform distribution on $(0,1)$, set $u=U_1$ and $v=(2U_2-1)\sigma \sqrt{2/e}$, and accept precisely whenReject and repeat otherwise. On acceptance return $X=v/u$. One can compare logarithms instead of exponentials to avoid underflow.

For a proof, the change of variables $(u,x)\mapsto (u,v=ux)$ has Jacobian determinant $u$. A uniform point in $\mathcal{R}$ therefore gives an $x$-density proportional toNormalizing yields the required normal distribution with mean zero and variance ${\sigma }^2$. The region area is $\frac{1}{2}\int h(x)dx=\sigma \sqrt{\pi /2}$; dividing it by the rectangle area $2\sigma \sqrt{2/e}$ gives acceptance probability $\sqrt{\pi e}/4$.

For $\alpha ,\beta >0$, integrating the Weibull distribution probability density function givesThe inverse transform sampling method takes $U$ uniform on $(0,1)$ and solves $G(X)=U$. HenceSince $1-U$ is also uniform, $\{-\beta \log U{\}}^{1/\alpha }$ is an equivalent implementation. For any $x\ge 0$, the resulting event $X\le x$ has probability $G(x)$, verifying its distribution.

Assume $0\le p\le 1$. Draw an independent component indicator with a Bernoulli distribution of success probability $p$. On success, generate the normal distribution from part (i) and return its absolute value; this has the half-normal distribution probability density function $2f(x)$ on $x\ge 0$. On failure, generate the Weibull distribution from part (ii). Use fresh independent uniforms for the chosen component.

The law of total probability gives the output probability density functionas required. The factor two belongs inside the half-normal component; the probability of choosing that component is $p$, not $2p$.